Substitution of Roots in a Polynomial (1 Viewer)

[murraysp]

ax^3 + bx^2 + cx + d
has roots alpha, beta, gamma and delta

find equation with
alpha^2...

i understand that use x^1/2 i can do this question

find equation with
1/alpha

from the examples i see you use 1/x

i dont get it why is it for one you rearrange and for the other you don't.

and can you also explain alpha + 1 equation, what do you sub?

Thanks

 

[khorne]

a+1 => x-1

 

[murraysp]

1. ok a+1=x-1

2. and a^2=sqrtx

3. but 1/a=1/x

how can that be?

i do not get the pattern

3. doesn't fit

 

[MOP777]

hey, hopefully this shows you why it is true.
the inv function of a+1 = a-1
inv function of x^2 = root(x)
and below is the inv. function of 1/x

 

[Rezen]

I dont think inverse functions is quite the reason why this works. if this was the case then if i wanted to find a polynomial with roots 1/(alpha^2), etc.. then i woould sub in sqrt(1/alpha). which leads to a polynomial fo degree four and hence is obviously wrong. (assuming my algebra wasn't shit.)

My understanding of this is that,

Hopefully i explained ok. ><

 

[Trebla]

I dont think inverse functions is quite the reason why this works. if this was the case then if i wanted to find a polynomial with roots 1/(alpha^2), etc.. then i woould sub in sqrt(1/alpha). which leads to a polynomial fo degree four and hence is obviously wrong. (assuming my algebra wasn't shit.)

If it is subbed into a polynomial of degree 3, then you wouldn't get a polynomial of degree 4, it would still be of degree 3 once you've rearranged it into a polynomial.

The inverse function idea is indeed the mechanism behind the substitution. In the example of roots 1/α, 1/β, ... etc notice that the inverse of a hyperbola y = 1/x is itself.
In general when you make the polnomial substitution of roots f(α), f(β), .... etc the approach is to
let y = f(x)
=> x = f^-1(y)
then sub it into the polynomial equation P(x) and rearrange to get a new polynomial in terms of y which then has roots f(α), f(β), etc...